Answer
$(a)\space 2P_{0}$
$(b)\space \frac{1}{2}(3+\sqrt 3)t_{0}$
Work Step by Step
(a) Given that, $P(t)=2P_{0}(1-\frac{(t-t_{0})^{2}}{t_{0}^{2}})$
Let's simplify this equation,
$P(t)=2P_{0}(1-\frac{(t^{2}-2t_{0}t+t_{0}^{2})}{t_{0}^{2}})$
$P(t)=2P_{0}(\frac{t_{0}^{2}-t^{2}+2t_{0}t-t_{0}^{2}}{t_{0}^{2}})=-2P_{0}(\frac{t^{2}-2t_{0}t}{t_{0}^{2}})-(1)$
Let's differentiate this equation by t & we get,
$\frac{d}{dt}P(t)=\frac{-2P_{0}}{t_{0}^{2}}(\frac{d}{dt}{t^{2}-2t_{0}\frac{d}{dt}t})$
$\frac{d}{dt}P(t)=\frac{-2P_{0}}{t_{0}^{2}}(2t-2t_{0})-(2) $
When P(T) in its maximum value,
$\frac{d}{dt}P(t)=0$
$(2)=\gt\space -(\frac{2P_{0}}{t_{0}^{2}})(2t-2t_{0})=0$
$2t-2t_{0}=0$
$t=t_{0}$
At this time the power of the second machine gets its maximum value.
Let's apply this value in equation (1)
$P(t_{0})=-2P_{0}(\frac{t_{0}^{2}-2t_{0}^{2}}{t_{0}^{2}})=\gt P(t_{0})=2P_{0}$
(b) To find the earliest time at which both machines have done the same amount of work, we have to intergrate equation (1) by t.
$\int P(t)\space dt=-(\frac{2P_{0}}{t_{0}^{2}})(\int t^{2}\space dt-2t_{0}\int t\space dt)$
$W(t) =-(\frac{2P_{0}}{t_{0}^{2}})(t^{2}/3-2t_{0}t^{2}/2)-(3)$
At that time worksdone by the machine 1 is $= P_{0}t-(4)$
$(3)=(4)=\gt$
$P_{0}t=-(\frac{2P_{0}}{t_{0}^{2}})(t^{2}/3-2t_{0}t^{2}/2)$
$t_{0}^{2}=-\frac{2}{3}t^{2}+2t_{0}t$
$3t_{0}^{2}=-{2}t^{2}+6t_{0}t$
$t^{2}-3t_{0}t+\frac{3}{2}t_{0}^{2}=0$
To find the value of t we use the equation of roots of quadratic equation
$t=\frac{-(-3t_{0})\pm\sqrt {9t_{0}^{2}-6t_{0}^{2}}}{2}=\frac{3t_{0}\pm t_{0}\sqrt 3}{2}$
$t=\frac{1}{2}(3-\sqrt 3)$
